MHTCET 16 May 2024 Shift 1 Paper

Concept:In a series combination of capacitors, the charge on each capacitor is the same and equals $C_{\mathrm{eq}}E$. Removing the dielectric changes the capacitance of B and hence the total stored charge.Explanation:Let the capacitance of capacitor A be $C$.With the dielectric in B, the capacitance of B becomes $KC$.For the series combination,$$\frac{1}{C_{\mathrm{eq}}} = \frac{1}{C} + \frac{1}{KC} = \frac{K+1}{KC}$$so,$$C_{\mathrm{eq}} = \frac{KC}{K+1}$$The initial charge on each capacitor is,$$Q_A = Q_B = C_{\mathrm{eq}}E = \frac{KCE}{K+1}$$After removing the dielectric slab, the capacitance of B becomes $C$ again.The new equivalent capacitance is,$$\frac{1}{C_{\mathrm{eq}}^{\prime}} = \frac{1}{C} + \frac{1}{C} = \frac{2}{C}$$$$C_{\mathrm{eq}}^{\prime} = \frac{C}{2}$$Hence, the new charge on each capacitor is,$$Q_A^{\prime} = Q_B^{\prime} = \frac{CE}{2}$$Comparing the charges on capacitor B,$$\frac{Q_B^{\prime}}{Q_B} = \frac{\frac{CE}{2}}{\frac{KCE}{K+1}} = \frac{K+1}{2K}$$The same ratio applies to capacitor A,$$\frac{Q_A^{\prime}}{Q_A} = \frac{K+1}{2K}$$Answer:Option D: $\frac{Q_B^{\prime}}{Q_B} = \frac{K+1}{2K}$